Câu 1:
a: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3x+9}{9-x}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-3x-9+2\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-3x-9+2x+6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
b: Thay \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) vào Q, ta được:
\(Q=\dfrac{3}{\sqrt{\left(\sqrt{3}-1\right)^2}-1}=\dfrac{3}{\sqrt{3}-1-1}\)
\(=\dfrac{3}{\sqrt{3}-2}=-3\left(2+\sqrt{3}\right)\)
c: Đặt A=Q:P
\(=\dfrac{3}{\sqrt{x}-1}:\dfrac{3}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)
Để A nguyên thì \(\sqrt{x}+3⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+4⋮\sqrt{x}-1\)
=>\(4⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{2;0;3;-1;5;-3\right\}\)
=>\(\sqrt{x}\in\left\{2;0;3;5\right\}\)
=>\(x\in\left\{0;4;9;25\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;4;25\right\}\)
